The Final Exam Schedule is Final!
Friday, December 18, 1:30 - 4:30 pm
Frank Kennedy Brown & Gold Gyms
The whole course
30 multiple choice questions
Formula sheet provided
Seating (this info is on Aurora)
Brown Gym: A - S
Gold Gym: T - Z
Monday, November 16, 2009
25
This Week: Experiment 4
Centripetal Force
WileyPLUS Assignment 4
Chapters 8, 9, 10
Due: Friday, November 27 at 11 pm
Next Week: Tutorial & Test 4
Monday, November 16, 2009
26
Variation of Pressure with Depth
For the column of fluid to be in equilibrium:
F1 = P1A
A
mg = F2 - F1 = P2A - P1A
As mass, m = volume x density = Ah!
Density, !
h
Ah!g = A(P2 - P1)
so P2 = P1 + !gh
A
F2 = P2A
The pressure is the same at all
points at the same depth
Monday, November 16, 2009
27
11.88/20: The blood pressure in the heart is 16 kPa. If an artery in
the brain is 0.45 m above the heart, what is the pressure in the
artery? Ignore any pressure changes due to blood flow.
P1 The brain
h = 0.45 m
P2 The heart
P2 = P1 + !gh,
so P1 = P2 – !gh
Density of blood = 1060 kg/m3
P1 = 16,000 – (1060 kg/m3)!(9.8 m/s2)!(0.45 m) = 11,325 Pa.
Note: P1, P2 are pressures above atmospheric pressure, the socalled “gauge pressure”. A tire pressure gauge measures gauge
pressure – zero on the gauge means the tire is flat and the
pressure inside the tire is atmospheric pressure.
Monday, November 16, 2009
28
Pumping water
from a well
Which method will pump
water from a deep well?
For the pipe at the right:
P1 = Patmos – !gh
H
P1
P1 is reduced to zero
(vacuum), when
h = Patmos/!g
= 101,300/(1000!9.8)
= 10.3 m
= maximum depth that
can be pumped from
above
h
Patmos
Monday, November 16, 2009
29
Mercury Barometer
P2 = Patmos = P1 + !gh
P1 = 0
Patmos = 0 + !gh
Column of mercury
!Hg = 13.6 × 103 kg/m3
Atmospheric pressure can be
measured in “mm of mercury”.
h
Standard atmospheric pressure:
101.3 × 103 Pa = !Hg gh
P2 = Patmos
101.3 × 103
h=
13.6 × 103 × 9.8
h = 0.76 m = 760 mm Hg
1 mm Hg = 1 Torr
Monday, November 16, 2009
30
Open Tube Manometer
Open to
atmosphere
P1 = Patm
Low density gas
P2
P2 = P1 + !gh
ρ
P2
= Patm + !gh
P1, P2, Patm are absolute pressures
Liquid, eg Hg
P2 – Patm = !gh = “gauge pressure” inside the box
Monday, November 16, 2009
31
A mercury manometer reads 747 mm on the roof of a building
and 760 mm on the ground. Assuming a constant value of 1.29
kg/m3 for the density of the air, find the height of the
building.
Pground = Proo f + !air gh
So, h =
Pground − Proo f
!air g
The atmospheric pressure increases by 13 mm Hg, that is, by:
13
13
atmospheres =
× 101.3 = 1.733 kPa
760
760
Therefore, h =
Monday, November 16, 2009
1733
= 137 m
1.29 × 9.8
32
11.99/29: A 1 m tall container is filled to the brim, part way
with mercury, the rest of the way with water. The container is
open to the atmosphere. What must be the depth of the
mercury so that the absolute pressure at the bottom is twice
atmospheric pressure?
P1 = 1 atmosphere
!1 = 1,000 kg/m3
!2 = 13,600 kg/m3
P2 = 2 atmospheres
y=1m
H2O
Hg
y=h
y=0
Monday, November 16, 2009
33
Pascal’s Principle
F1
!P =
A1
P + !P
Monday, November 16, 2009
∆P =
F2
A2
P + !P
h
P + !gh + "P
Increase the pressure
at the left by applying
force F1 on a piston.
Pascal: the same pressure
increase "P is communicated
everywhere in the enclosed
fluid.
h
P + !gh + "P
34
Pascal’s Principle
Apply a small force F1
at the left:
!P =
F2
F1
A1
!P =
F1 F2
=
A1 A2
A2
The force is multiplied:
F2 = F1 ×
A2
A1
Monday, November 16, 2009
35
Hydraulic jack
The input piston (on the left)
has a radius r1 = 0.012 m and at
the right r2 = 0.15 m. The
combined weight of car and
plunger is 20,500 N. The oil in
the jack has density 800 kg/m3.
20,500 N
PA = PB
A1
A2
(a) What force F1 is needed to support the car and output plunger
when the bottom surfaces of piston and plunger are at the same
height? (h = 0)
Monday, November 16, 2009
36
Hydraulic jack
(b) When h = 1.1 m?
20,500 N
PA = PB
A1
A2
Monday, November 16, 2009
37
11.38/35: A spring (k = 1600 N/m) is attached to the left piston, a
40 kg rock is placed on the right. By how much is the spring
compressed?
Left and right pistons are
at about the same height.
So: ∆P =
mg
F
=
65 cm2
15 cm2
F = 40g ×
15
= 90 N
65
F
mg
Spring is compressed by:
x=
F
90
=
k 1600
x = 0.056 m
Monday, November 16, 2009
38
Archimedes’ Principle
Consider the equilibrium of the column of
fluid. It experiences pressure forces from
above and below.
The net upward force (buoyant force) on
the column is:
Fb = (P2 − P1)A
Fb
and P2 − P1 = !gh
increase of pressure
with depth
So, Fb = !hAg = !V g
!V = mass of fluid that occupied the volume V.
Fluid of density !
Buoyant force = weight of displaced fluid
Monday, November 16, 2009
39
Archimedes’ Principle
Buoyancy force = !Vg = weight of fluid
displaced by the object
Floating
object
Fb
Volume immersed
=V
Fluid, density !
Fb
Fluid of density !
mg
The buoyant force acting on an object
partially or completely immersed in a fluid
is the weight of the fluid that is displaced.
For a floating object, Fb = mg
That is, !Vg = mg
Monday, November 16, 2009
40
Clickers!
You put an ice cube in a glass and fill the glass with water. As the
ice cube melts, does the water level drop, stay the same, or rise?
Fb
Ice cube
Fb = mg
Volume immersed
=V
mg
A) The level of water rises
B) The level of water falls
C) The level of water remains the same
The ice cube floats by displacing its own weight of liquid water.
So, when the ice cube melts, it occupies a volume equal to that of
the liquid water it displaced when it was floating...
That is, the ice cube contracts into the volume coloured green...
Monday, November 16, 2009
41
The Goodyear blimp contains 5400 m3 of helium of density
0.179 kg/m3. Find the weight of the load WL that the airship can
carry if the density of the air is 1.2 kg/m3.
Fb
He
V
Air
WL + WHe
To float in the air: Fb = WL + WHe
Monday, November 16, 2009
42
Fluids, the story so far...
Density = Mass/Volume
Specific gravity =
Density of substance
Density of water at 4◦C
Pressure = Force/Area
Increase of pressure with depth:
P2 = P1 + !gh
Pascal’s Principle: pressure changes are communicated uniformly
throughout an enclosed fluid
Archimedes’ Principle: the buoyant force is equal to the weight of
displaced fluid
Monday, November 16, 2009
43
11.43/38: A duck is floating on a lake with 25% of its volume
beneath the water. What is the average density of the duck?
Fb = buoyant force
0.25V
Surface of water
mduck g = weight of duck, total volume = V
The duck floats, so the weight of the duck is equal to the weight
of water that has 25% of the volume of the duck.
Monday, November 16, 2009
44
11.47/44: A solid object has an apparent weight of 15.2 N when
completely immersed in ethyl alcohol.
When completely submerged in water, its apparent weight is 13.7 N.
What is the volume of the object?
Density of ethyl alcohol = 806 kg/m3.
T
Fb
Weigh the object by supporting it by a string and
measuring the tension in the string.
In ethyl alcohol: T1 = mg − Fb1 = 15.2 N
mg
In water:
T2 = mg − Fb2 = 13.7 N
Monday, November 16, 2009
45
Fluids, continued...
Pascal’s Principle: a change in pressure is transmitted
equally throughout an enclosed fluid.
Archimedes’ Principle: the buoyant force acting on an
object is equal to the weight of fluid displaced by the
object.
Monday, November 16, 2009
46